A math professor of mine was recommending that I go over integration techniques before doing courses involving differential equations. Since I plan on doing at least 3 such courses in the fall, (comp. mechanics, classical mechanics, and diff. eq. itself) I need to catch up on this stuff.

 

So, I was doing some integration by parts practice problems, and I've run into trouble already.

 

One such problem was to find the indefinite integral of LN(2x+1)dx with respect to x. So I set up u = LN(2x+1) and dv = dx, implying v = x and du = 2/(2x+1)

 

Using the formula INT(u)dv = uv-INT(v)du, this yielded LN(2x+1)-INT(2/(2x+1))dx

 

So for the integral within an integral, I used parts again. u = 1/(2x+1) and dv = 2dx implies du = -2dx/((2x+1)^2) and v = 2x

 

Using the same formula, this yielded (2x)/(2x+1)-INT((2xdx)/(4x^2+4x+1))

 

So I tried a w-substitution. If w=4x^2+4x+1, then dw=(8x+4)dx; if it were just 8xdx I could substitute it, but that +4 part makes it seem like it would be mathematically improper. What did I do wrong?

One such problem was to find the indefinite integral of LN(2x+1)dx with respect to x. So I set up u = LN(2x+1) and dv = dx, implying v = x and du = 2/(2x+1)

Correction: [math]du = \frac 2{2x+1} dx[/math]. You forgot the dx.

 

Using the formula INT(u)dv = uv-INT(v)du, this yielded LN(2x+1)-INT(2/(2x+1))dx

Look at what you wrote for u, v, and du (with my correction). Do you see the error? Once you make a mistake in an intermediate step all work from that point on is no good.

Correction: [math]du = \frac 2{2x+1} dx[/math]. You forgot the dx.

 

 

Look at what you wrote for u, v, and du (with my correction). Do you see the error? Once you make a mistake in an intermediate step all work from that point on is no good.

I'm guessing it was that my initial uv term was supposed to be x*Ln(2x+1) as opposed to just Ln(2x+1) itself?

That's one error. What about the [math]\int v\,du[/math] term?

A math professor of mine was recommending that I go over integration techniques before doing courses involving differential equations. Since I plan on doing at least 3 such courses in the fall, (comp. mechanics, classical mechanics, and diff. eq. itself) I need to catch up on this stuff.

 

So, I was doing some integration by parts practice problems, and I've run into trouble already.

 

One such problem was to find the indefinite integral of LN(2x+1)dx with respect to x. So I set up u = LN(2x+1) and dv = dx, implying v = x and du = 2/(2x+1)

 

Using the formula INT(u)dv = uv-INT(v)du, this yielded LN(2x+1)-INT(2/(2x+1))dx

 

So for the integral within an integral, I used parts again. u = 1/(2x+1) and dv = 2dx implies du = -2dx/((2x+1)^2) and v = 2x

 

Using the same formula, this yielded (2x)/(2x+1)-INT((2xdx)/(4x^2+4x+1))

 

So I tried a w-substitution. If w=4x^2+4x+1, then dw=(8x+4)dx; if it were just 8xdx I could substitute it, but that +4 part makes it seem like it would be mathematically improper. What did I do wrong?

 

You have started it right.

let, u = ln(2x+1), du = 2.dx/(2x+1)

dv = dx, v = x

 

now the integral will become,

 

x.ln(2x+1) - int(2x.dx/(2x+1))

 

by algebraic manupulation, it will turn out to be,

 

x.ln(2x+1) - int(dx) + int(dx/(2x+1))

 

which will yield:

 

x.ln(2x+1) - x + ln(2x+1) /2

 

x.ln(2x+1) - x + ln(2x+1) /2

 

Verify your result by differentiating again.